StatementSwimsuit development took massive leap forward in 2009 with the introduction of 'Speedo LZR' suits with polyurethane suits such as the 'Aren...

This problem is much more ambitious than the usual problems, in the sense that it requires putting together a greater number of parts. But if you tackle the various parts as suggested, you should find that they are not, individually, especially difficult, and the problem as a whole exemplifies the power of the energy-conservation method for analyzing oscillation problems. You are no doubt familiar with the phenomenon of water sloshing about in the bathtub. The simplest motion is, to some approximation, one in which the water surface just tilts as shown but seems to remain more or less flat. A similar phenomenon occurs in lakes and is called a seiche (pronounced: saysh). Imagine a lake of rectangular cross section, as shown, of length $L$ and with water depth $h(\ll L) .$ The problem resembles that of the simple pendulum, in that the kinetic energy is almost entirely due to horizontal flow of the water, whereas the potential energy depends on the very small change of vertical level. Here is a program for calculating, approximately, the period of the oscillations: (a) Imagine that at some instant the water level at the extreme ends is at $\pm y_{0}$ with respect to the normal level. Show that the increased gravitational potential energy of the whole mass of water is given by $U=\frac{1}{6} b \rho g L y_{0}^{2}$ where $b$ is the width of the lake. You get this result by finding the increased potential energy of a stice a distance $x$ from the center and integrating. (b) Assuming that the water flow is predominantly horizontal, its speed $v$ must vary with $x$, being greatest at $x=0$ and zero at $x=\pm L / 2 .$ Because water is incompressible (more or less) we can relate the difference of flow velocities at $x$ and $x+d x$ to the rate of change $d y / d t$ of the height of the water surface at $x$. This is a continuity condition. Water flows in at $x$ at the rate $c h b$ and flows out at $x+d x$ at the rate $(v+d v) h b$. (We are assuming $y_{0} \ll h$.) The difference must be equal to $(b d x)(d y / d t)$, which represents the rate of increase of the volume of water contained between $x$ and $x+d x$. Using this condition, show that $$ v(x)=c(0)-\frac{1}{h L} x^{2} \frac{d y_{0}}{d t} $$ where $$ v(0)=\frac{L}{4 h} \frac{d y_{0}}{d t} $$ (c) Hence show that at any given instant, the total kinetic energy associated with horizontal motion of the water is given by $$ \boldsymbol{K}=\frac{1}{60} \frac{b \rho L^{3}}{h}\left(\frac{d y_{0}}{d t}\right)^{2} $$ To get this result, one must take the kinetic energy of the slice of water lying between $x$ and $x+d x$ (with volume equal to $b h d x$ ), which moves with speed $v(x)$, and integrate between the limits $x=\pm L / 2$. (d) Now put $$ \boldsymbol{K}+\boldsymbol{U}=\text { const. } $$ This is an equation of the form $$ A\left(\frac{d y_{0}}{d t}\right)^{2}+B y_{0}^{2}=\text { const. } $$ and defines SHM of a certain period. You will find that this period depends only on the length $L$, the depth $h$, and $g$. [Note: This theory is not really correct. The water surface is actually a piece of a sine wave, not a plane surface. But our formula is correct to better than $1 \%$. (The true answer is $T=2 L / \sqrt{g h})$.] (e) The Lake of Geneva can be approximated as a rectangular tank of water of length about $70 \mathrm{~km}$ and of mean depth about $150 \mathrm{~m}$. The period of its seiche has been observed to be about $73 \mathrm{~min}$. Compare this with your formula.

So this problem is all about Samantha kayaking and so she drops her um life preserver overboard when she launches and that's the keys when she launched the boat. So after she paddles upstream a bit, she has to go back to get it. So question A wants vectors represent her upstream displacement from the launch to where she turned around and her downstream displacement to get the life jacket. So that's why A is showing her upstream displacement is shorter than her downstream displacement. That's because she has to go past the launch point in order to get that life jacket because it has been floating downstream with the water. So for be it wants to know what the overall displacement of her paddling and the life jacket are while they're the same. It turns out that from where she launched which is where she left the lifejacket, Both the boat and um the life jacket moved the same displacement downstream. So if we look at a, we have our upstream and downstream and if we add those two things together, the entirety of the upstream paddling cancels with most of the downstream paddling. And we're just left with this short displacement, which is the same for both the kayak and for the life jacket. Now, see is asking for the reference frame of the water. So from the reference frame of the water which is moving downstream, um What about Samantha's displacement relative to the water? Well, it's zero. It turns out as she's moving upstream. The water is moving downstream when she turns around, she is then moving downstream with the water. But the key is when she drops that life jacket in the water at the launch, that water in the life jacket travel together down word, that means when she turns around and gets back to that life jacket, she is back to where that water. Um Exactly that water that she started at. So it's like if you were to imagine that where the life jacket is in the water is zero, even though it's technically moving in the frame of reference of the riverbank in the frame of reference of the water, it's the same place. And so her displacement is zero in the reference frame of the water. She comes back to her. She started question d asked does she spend more time paddling upstream, paddling downstream or the same each way? It should be the same. And the reasoning is because of our equation speed equals distance over time. So we know from the question that her speeds relative to the water are the same upstream and downstream. Now she had to go the same distance relative to the water in both directions as we explained in part C. So that means that the time has to be the same if it's the same speed.